Homework Question -Vector Calculus Area

Question

I want to calculate the area of a semi-circle. I can use this $\iint x^2+y^2\,\mathrm dx\,\mathrm dy$ or I can use this $\iint r^2 r^2 \sin(\phi) \,\mathrm \,d\theta\,\mathrm dr$. I can see where the $r\,\mathrm d\theta \, dr$ comes from but why is it wrong to say $x=r\cos\theta$ and $y=r\sin \theta$, then differentiate and multiply and get $$(dr)^2\cos\theta \sin\theta +r\,dr\,d\theta \cos^2 \theta −r\,dr\,d \theta \sin^2\theta −r^2(d\theta)^2\sin\theta \cos\theta?$$

Obviously I can't use the above in the integral but I cant see where this is wrong.

Answer 1

Firstly, you have the wrong formula. The area of the semicircle in Cartesian coordinates would be

$$\int_{-r}^r \sqrt{r^2-x^2} \ dx$$

or, as a double integral,

$$\int_{-r}^r \int_0^{\sqrt{r^2-x^2}} \,dy \ dx$$

Secondly, you're not following the process for changing variables in double integrals, given on the linked source as $$\iint\limits_R f(x,y) \, dx \, dy = \iint\limits_S f[x(u,v),y(u,v)] \left| \frac{\partial(x,y)}{\partial(u,v)} \right| \, du\,dv$$

which ultimately leads to your confusion.

When you change variables in double integrals, you need to compute the Jacobian of your transformation. That means that when you substitute $x=r\cos \theta$ and $y=r\sin \theta$, you need to compute

$$\left| \frac{ \partial(x,y)}{ \partial(r, \theta)} \right| =\left| \begin{matrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{matrix} \right|$$

In this case, that gives you

$$\left| \begin{matrix} \cos \theta & -r\sin\theta \\ \sin \theta & r\cos\theta \end{matrix} \right| =r(\cos^2\theta+\sin^2\theta)=r$$

Once you have this, you can safely convert the integral into polar coordinates to get

$$\int_{-r}^r \int_0^{\sqrt{r^2-x^2}}dy \ dx=\int_0^\pi \int_{0}^r r \ dr \ d\theta=\frac{\pi r^2}{2}$$

Answer 2

A product of differentials is anti-symmetric. That is, $dr\,d\theta= -d\theta \,dr$. It is also follows that $dr\,dr= -dr\,dr$ so that $dr\,dr= 0$ and $d\theta \,d\theta= -d\theta \,d\theta$ so that $d\theta \,d\theta= 0$.

If $x= r\cos(\theta)$ and $y= r\sin(\theta)$ then $dx= \cos(\theta)\,dr- r \sin(\theta)\,d\theta$ and $dy= \sin(\theta)\,dr+ r \cos(\theta)\,d\theta$ so that

\begin{align} dx\,dy = {} & (\cos(\theta)\,dr- r\sin(\theta)\,d\theta)(\sin(\theta)\,dr+ r \cos(\theta)\,d\theta) \\[8pt] = {} & \cos(\theta)\sin(\theta)\,dr\,dr+ r\cos^2(\theta) \, dr \, d\theta \\ & {} - r \sin^2(\theta)\,d\theta \,dr- r^2 \sin(\theta) \cos(\theta) \,d\theta \,d\theta \\[8pt] = {} & 0+ r \cos^2(\theta)\,dr \,d\theta+ r \sin^2 \,dr\,d\theta+ 0 \\[8pt] = {} & r \,dr \,d\theta \end{align}